Be $\Sigma_{\lambda}$ the subset in $\mathbb{CP}(2)$ described in affine coordinates $(1,y,z)$ by the zero loci of the function $$y^2-z(z-1)(z-\lambda)$$. We can compactify it adding $(0,1,0)$.
We then obtain a torus, represented as two sheeet attached by two branch cuts, one ending in the point at infinity (0,1,0).
Each sheet is described by a coordinate $z$, via $z \to (1,y(z),z)$, where a choice of sign has been made for $y(z) = \pm \sqrt{z (z-1) (z-\lambda)}$. Let's say $+$ for the upper sheet and $-$ for the lower sheet.
Now, be $(\sigma,y_\sigma)$ be a point on this torus. It is implied that in $y_\sigma$ a choice of sign has been made. We can consider the differential form $\frac{dz}{y} \frac{y+y_\sigma}{z-\sigma}$.
This is a meromorphic differential form, and it seems to me that it has a single pole: (\sigma,y_\sigma). Note that $(\sigma,-y_\sigma)$ is not a pole due to the factor $(y+y_a)$ which cancel $(z-\sigma)$.
The residue at this pole is $1/2$.
However, the residue theorem for Riemann surfaces states that the sum of residues of a differential form is zero. In fact, it should be impossible to find a form with a single pole. So...where's it the problem with my form? Am I missing another pole?
EDIT: The point at infinity is the other pole, with opposite residue.
Closely related question: Since $deg(\omega)=0$ for any differential form on a torus, there should also be two zeros, or a single zero with double multiplicity. Any clue where this could be?